This question evaluates the connection between differentiability and continuity for piecewise functions. For this f(x), at x=0, the left-hand limit is 0 but f(0)=1, creating a discontinuity. Since differentiability implies continuity, the lack of continuity means f'(0) does not exist. The pieces may be differentiable individually, but the jump at x=0 breaks continuity. Choice C is a tempting distractor, claiming continuity at x=0 ensures the derivative exists, but it fails because the function is actually discontinuous there. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, considering redefined points in otherwise continuous functions. For this f(x), lim x→0 sin x = 0, but f(0)=1, creating a discontinuity. Differentiability requires continuity, so f'(0) cannot exist here. The redefinition breaks the necessary limit agreement. A tempting distractor might say yes because redefining one point doesn't affect differentiability, but it does if it causes discontinuity. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question examines the link between differentiability and continuity, specifically regarding vertical tangents. A vertical tangent at x=5 indicates that the slope approaches infinity, meaning the limit of the difference quotient does not exist as a finite number. For p'(5) to exist, this limit must be finite and equal from both sides, but an infinite slope prevents that. The function can be continuous at x=5 with a vertical tangent, yet differentiability requires a finite derivative. Choice C is a tempting distractor, stating continuity ensures the derivative exists, but it fails because vertical tangents make the derivative undefined despite continuity. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, analyzing piecewise linear functions. For this f(x), it is continuous at x=0 with value 2 matching both limits. However, the left-hand derivative is 1, while the right-hand is -1, creating a corner. This difference means f'(0) does not exist. A tempting distractor might say yes because of continuity, but slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity for the absolute value function. For f(x) = |x|, at x=0, the left-hand derivative is -1 and the right-hand is 1, which are unequal. This mismatch means there is no unique derivative at x=0. The function is continuous at x=0, but the corner prevents differentiability. Choice A is a tempting distractor, claiming continuity implies differentiability, but it fails because of the differing slopes. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, emphasizing that while continuity is necessary for differentiability, it is not sufficient without additional conditions. The function is given as continuous at x=9, which satisfies the prerequisite for differentiability. Additionally, the one-sided derivatives are equal, meaning the left-hand derivative matches the right-hand derivative at x=9. Therefore, the two-sided derivative exists at x=9, confirming differentiability. A tempting distractor might suggest that matching function values alone guarantee differentiability, but this ignores the need for matching slopes from both sides. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question tests differentiability versus continuity in piecewise functions with jumps. For this f(x), at x=1, the left-hand limit is 1 and the right-hand is 3, creating a jump discontinuity. Without continuity, f'(1) cannot exist, regardless of the pieces being differentiable elsewhere. The added constant causes the discontinuity. Choice C is a tempting distractor, claiming a jump still allows a derivative, but it fails because continuity is required. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, stressing the need for function definition and continuity. The function f(x) = |x|/x is undefined at x=0, violating the basic requirement for differentiability. Without a function value at x=0, continuity cannot hold, and the derivative limit cannot be computed. Therefore, f'(0) does not exist. A tempting distractor might suggest yes if one-sided derivatives match, but the lack of definition overrides this. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, identifying features that allow continuity without differentiability. A corner at x=2 maintains continuity but causes unequal one-sided derivatives. This prevents differentiability while preserving continuity. Other options like jumps or holes would break continuity. A tempting distractor might be a hole, but holes cause discontinuity, not fitting the condition. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

This question assesses the connection between differentiability and continuity in AP Calculus AB, evaluating combined absolute value functions. For f(x) = |x-4| + $|x-4|^2$, it is continuous at x=4 with value 0. However, the left-hand derivative is -1, while the right-hand is 1, due to the |x-4| term. This mismatch creates a corner, so f'(4) does not exist. A tempting distractor might say yes because of continuity, but the slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.